Question

Write down the Lagrangian for a projectile (subject to no air resistance) in terms of its Cartesian coordinates \((x, y, z),\) with \(z\) measured vertically upward. Find the three Lagrange equations and show that they are exactly what you would expect for the equations of motion.

Short answer

The Lagrangian is \( L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - mgz \). The equations: \( \ddot{x} = 0 \), \( \ddot{y} = 0 \), \( \ddot{z} = -g \).

Step-by-step solution

Understanding the Lagrangian Function

The Lagrangian function, defined as \( L = T - V \), where \( T \) is the kinetic energy and \( V \) is the potential energy, is used to derive the equations of motion. For a projectile in three-dimensional space, we'll express both the kinetic and potential energy in terms of the Cartesian coordinates \((x, y, z)\).

Express Kinetic Energy in Cartesian Coordinates

The kinetic energy \( T \) of a projectile with mass \( m \) moving with velocity components \( \dot{x}, \dot{y}, \dot{z} \) is given by \( T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) \). These are the squared velocity components in the Cartesian system.

Express Potential Energy in Cartesian Coordinates

The potential energy \( V \) of a projectile in a gravitational field (with constant gravity \( g \)) depends on the vertical position \( z \). It is expressed as \( V = mgz \), where \( z \) is the height above the reference point.

Write the Lagrangian

Substitute the expressions for kinetic and potential energy into the Lagrangian: \[ L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - mgz \] This function represents the Lagrangian of the system.

Derive the Lagrange Equations

Using the Euler-Lagrange equation, \( \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0 \), apply it to each coordinate \( x, y, \) and \( z \).

Equation for the x-direction

Applying the Lagrange equation to \( x \):\[ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{x}} \right) = m\ddot{x} \quad \text{and} \quad \frac{\partial L}{\partial x} = 0 \]Thus, the equation becomes: \[ m\ddot{x} = 0 \] indicating no acceleration in the \( x \)-direction.

Equation for the y-direction

Similarly, apply the Lagrange equation to \( y \):\[ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{y}} \right) = m\ddot{y} \quad \text{and} \quad \frac{\partial L}{\partial y} = 0 \]Thus, \[ m\ddot{y} = 0 \] indicating no acceleration in the \( y \)-direction as well.

Equation for the z-direction

For the \( z \) coordinate, we have:\[ \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{z}} \right) = m\ddot{z} \quad \text{and} \quad \frac{\partial L}{\partial z} = -mg \]This results in:\[ m\ddot{z} = -mg \] which represents the acceleration due to gravity.

Summary of Equations of Motion

The Lagrange equations lead to the expected equations of motion: \[ \ddot{x} = 0, \quad \ddot{y} = 0, \quad \ddot{z} = -g \] These equations show that the projectile moves with constant velocity in the \( x \) and \( y \) directions, and with constant acceleration \( -g \) in the \( z \) direction.

Key concepts

Projectile Motion

Projectile motion describes the path of an object thrown into space, subject to gravitational forces alone. In our exercise, we're analyzing this motion without air resistance, focusing on three-dimensional space with coordinates

• \( (x, y, z) \)

Here, \( z \) is vertical. The key factors in projectile motion are:

• Initial Velocity: Determines how the object starts its trajectory.
• Gravity: Acts downward, affecting vertical motion.

In our case, the Lagrange equations reveal that horizontal movements (\( x \) and \( y \)) have no acceleration. The vertical motion (\( z \)) accelerates down due to gravity (-g). This analysis gives us an understanding of how an object like a ball thrown from a cliff behaves under gravity's influence.

Euler-Lagrange Equation

The Euler-Lagrange equation is a fundamental principle to derive equations of motion in Lagrangian mechanics. It connects the Lagrangian function to physical trajectories of systems. The equation is:\[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0\]Where \( L = T - V \), it represents the difference between kinetic and potential energy. Here’s how it applies:

• For each coordinate (\( x, y, z \)): We take partial derivatives related to velocity and position.
• Motion Derivation: This leads to equations describing how the system evolves over time.

In our projectile exercise, applying this to \( x \) and \( y \) showed no force acting horizontally thus no acceleration (\( \ddot{x} = 0 \)). But for \( z \), gravity force exists, leading to \( \ddot{z} = -g \), reflecting downward acceleration. Understanding the Euler-Lagrange equation lets us derive these natural laws from simple principles.

Kinetic and Potential Energy

Kinetic and potential energies are key to understanding Lagrangian mechanics. They are components of the Lagrangian function \( L = T - V \), representing how energy changes in motion.

Kinetic Energy (T)

Kinetic energy is associated with the motion of the projectile. Given by:\[ T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) \]

• Mass \( m \): How heavy the object is.
• Velocity Components (\( \dot{x}, \dot{y}, \dot{z} \)): Speed in the three axes.

These factors combine to show how fast an object is moving in the Cartesian plane.

Potential Energy (V)

Potential energy relates to the object's position in a gravitational field:\[ V = mgz\]

• Gravitational Force (\( g \)): Constant pulling the object down.
• Height (\( z \)): Position elevated above a reference point.

Thus, an object's potential energy changes as it moves vertically. Understanding these energies helps to reflect the trade-offs in motion, vital for predicting the projectile's path.